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#1 2009-08-28 06:28:15

JaneFairfax
Legendary Member

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Number of roots of polynomials




This can be proved using the equivalent of the division algorithm for polynomials over a field.





A simple proof of this using induction on the degree of
can be found on p.28 of D.J.H. Garling, A Course in Galois Theory (1986), Cambridge University Press.

Note that this applies to polynomials over a field only. This won’t work with rings in general, e.g.
and
in
. However, if
is an integral domain, we can apply the division algorithm to polynomials over its field of fractions
.

Let
be a polynomial of degree
in
. Let
be a root of
Considering
as a polynomial in
, we can write
where
and
. Then
.

Thus
. If
is another root of
distinct from
, write
where
and
. Then
as
.

Thus
. Continuing this way, we find that if
are all the distinct roots of
, then we have
for some
. It follows that
cannot have more than
roots in
since
has degree
and so cannot be factorized into more than
linear factors in
. And since a root of polynomial in
is also a root of a polynomial in
(Gauss’s lemma), our proof is complete. smile

If
is not an integral domain, it may be possible for an
th-degree polynomial in
to have more than
roots. For instance, as a polynomial over
, the quadratic polynomial
actually has 4 roots, namely 1, 3, 5, 7. That’s because
is not an integral domain and so we cannot construct its field of fractions.


Q: Who wrote the novels Mrs Dalloway and To the Lighthouse?

A: Click here for answer.
 

#2 2009-08-29 09:49:08

bobbym
Administrator

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Re: Number of roots of polynomials

Hi Jane;

Thanks for the reason why  x^2 + 7=0 mod 8 can have 4 solutions. I didn't know that.

Last edited by bobbym (2009-08-29 10:19:02)


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

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